| If \[{{x}^{2}}=y+z,\]\[{{y}^{2}}=z+x\]and \[{{z}^{2}}=x+y,\]then the value of \[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1},\]is [SSC (CPO) 2013, (10+2) 2011] |
A) \[-\,\,1\]
B) \[1\]
C) \[2\]
D) \[4\]
Correct Answer: B
Solution :
| Given, \[{{x}^{2}}=y+z\] |
| On adding x both sides, we get |
| \[{{x}^{2}}+x=x+y+z\] |
| \[\Rightarrow \] \[x\,\,(x+1)=x+y+z\] |
| \[\Rightarrow \] \[\frac{1}{x+1}=\frac{x}{x+y+z}\] (i) |
| Similarly, \[\frac{1}{y+1}=\frac{y}{x+y+z}\] ... (ii) |
| and \[\frac{1}{z+1}=\frac{z}{x+y+z}\] ... (iii) |
| On addition Eqs. (i), (ii) and (iii), we get |
| \[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\] \[=\frac{x+y+z}{x+y+z}=1\] |
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